2. What is the pH of buffer when 0.01 mol of NaOH is added to the...

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Answer 1

Answer #1

Answer

2.

Given, NH₃= 0.3 M, NH₄Cl= 0.2 M, pKa of NH₃= 9.25

the buffer of NH₃ and NH₄Cl will act as a basic buffer

pOH = pka + log [salt/acid]

pOH = 9.25 + log [NH₄Cl/ NH₃]

= 9.25 + log [0.2/0.3]

pH of buffer= 9.25 - 0.176= 9.074

pH of buffer= 14- pOH= 14- 9.074= 4.926

if we added 0.01/0.5= 0.02 M of NaOH (molarity (M) = moles/volume)

[NH₃]= 0.3 + 0.02= 0.32 M

[NH₄⁺]= 0.20 + 0.02= 0.18

pOH= 9.25 + log[0.18/0.32]

pOH= 9.25 - 0.250= 9

then pH will be = 14- 9= 5

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